Pascals Law

Relationship Between Depth and Pressure - h1012v3_28

Control Volume

Thermodynamics Heat Transfer and Fluid Flow Volume 3 of 3
Page Navigation
10

Fluid Flow CONTINUITY EQUATION (b) Pressure Force Area Force (Pressure) (Area) Area pr 2 F 17.2 lbf in² p 10 ft 2 144 in² 1 ft² 7.78 x 10⁵ lbf Pascal’s Law The pressure of the liquids in each of the previously cited cases has been due to the weight of the liquid. Liquid pressures may also result from application of external forces on the liquid. Consider the following examples. Figure 2 represents a container completely filled with liquid. A, B, C, D, and E represent pistons of equal cross-sectional areas fitted into the walls of the vessel. There will be forces acting on the pistons C, D, and E due to the pressures caused by the different depths of the liquid. Assume that the forces on the pistons due to the pressure caused by the weight of the liquid are as follows: A = 0 lbf, B = 0 lbf, C = 10 lbf, D = 30 lbf, and E = 25 lbf. Now let an external force of 50 lbf be applied to piston A. This external force will cause the pressure at all points in the container to increase by the same amount. Since the pistons all have the same cross-sectional area, the increase in pressure will result in the forces on the pistons all increasing by 50 lbf. So if an external force of 50 lbf is applied to piston A, the force exerted by the fluid on the other pistons will now be as follows: B = 50 lbf, C = 60 lbf, D = 80 lbf, and E = 75 lbf. This effect of an external force on a confined fluid was first stated by Pascal in 1653. Pressure applied to a confined fluid is transmitted undiminished throughout the confining vessel of the system. Rev. 0 Page 7 HT-03

Privacy Statement
Press Release
Contact